Question

a rectangular piece of metal is 25 in longer than it is wide. squares with sides 5 in long are cut from the four corners and the flaps are folded upward to form an open box. if the volume of the box is 1980 in³, what were the original dimensions of the piece of metal? what is the original width?

Explanation:

Step1: Let the original width be \(x\) inches.

The original length is \(x + 25\) inches. After cutting 5 - inch squares from each corner and folding up, the width of the box is \(x-2\times5=x - 10\) inches, the length of the box is \((x + 25)-2\times5=x + 15\) inches, and the height of the box is 5 inches.

Step2: Set up the volume equation.

The volume \(V\) of a rectangular - box is \(V=\text{length}\times\text{width}\times\text{height}\). We know \(V = 1980\) cubic inches. So \(5(x + 15)(x - 10)=1980\).
First, divide both sides by 5: \((x + 15)(x - 10)=396\).
Expand the left - hand side using the FOIL method: \(x^{2}-10x+15x-150 = 396\).
Combine like terms: \(x^{2}+5x-150 = 396\).
Rearrange to get a quadratic equation: \(x^{2}+5x-546 = 0\).

Step3: Solve the quadratic equation.

For a quadratic equation \(ax^{2}+bx + c = 0\) (\(a = 1\), \(b = 5\), \(c=-546\)), we can use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or factor.
Factor the quadratic equation: \(x^{2}+5x - 546=(x + 26)(x - 21)=0\).
Set each factor equal to zero: \(x+26 = 0\) gives \(x=-26\) (rejected since the dimension cannot be negative), and \(x - 21=0\) gives \(x = 21\).

Step4: Find the original length and width.

The original width \(x = 21\) inches.
The original length is \(x + 25=21 + 25 = 46\) inches.

Answer:

The original width is 21 inches and the original length is 46 inches.